2017-2018 ACM-ICPC Latin American Regional Programming Contest K-Keep it covered

题目描述

http://codeforces.com/gym/101889/attachments/download/7471/statements-2017-latam-regional.pdf

简要题意:

Solution

我们将红色格子看成度数为 $1$,白色格子看成度数为 $2$

然后每个点向周围点连边,求最大匹配是否等于点的度数和即可

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
#include <iostream>
#include <cstdio>
#include <queue>
#define maxm 21
#define maxn 810
#define INF 1000000000
using namespace std;

int dx[] = { 0, 0, 1, -1 };
int dy[] = { 1, -1, 0, 0 };

int n, m, id[maxm][maxm], s, t;

inline int D(int x, int y) { return (x - 1) * m + y; }

struct Edge {
int to, next, w;
} e[1000000]; int c1, head[maxn];
inline void add_edge(int u, int v, int w) {
e[c1].to = v; e[c1].w = w;
e[c1].next = head[u]; head[u] = c1++;
}

inline void Add_edge(int u, int v, int w) {
add_edge(u, v, w); add_edge(v, u, 0);
}

int dep[maxn];
bool bfs() {
fill(dep, dep + maxn, 0); dep[s] = 1;
queue<int> Q; Q.push(s);
while (!Q.empty()) {
int u = Q.front(); Q.pop();
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (!dep[v] && w > 0) {
dep[v] = dep[u] + 1;
if (v == t) return 1; Q.push(v);
}
}
}
return 0;
}

int dfs(int u, int _w) {
if (u == t || !_w) return _w;
int s = 0;
for (int i = head[u]; ~i; i = e[i].next) {
int v = e[i].to, w = e[i].w;
if (dep[v] == dep[u] + 1 && w > 0) {
int di = dfs(v, min(_w - s, w));
e[i].w -= di; e[i ^ 1].w += di;
s += di; if (s == _w) break;
}
}
if (s < _w) dep[u] = 0;
return s;
}

int dinic() {
int mf = 0;
while (bfs()) mf += dfs(s, INF);
return mf;
}

int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr); cout.tie(nullptr);

cin >> n >> m; s = 0; t = 2 * n * m + 1;
for (int i = 1; i <= n; ++i) {
char s[maxm]; cin >> s + 1;
for (int j = 1; j <= m; ++j)
id[i][j] = s[j] != 'o';
} int ans = 0;
for (int i = 1; i <= n; ++i)
for (int j = 1; j <= m; ++j) {
ans += id[i][j] + 1;
for (int k = 0; k <= id[i][j]; ++k)
if (i + j & 1) Add_edge(s, D(i, j) + k * n * m, 1);
else Add_edge(D(i, j) + k * n * m, t, 1);
for (int k = 0; k < 4; ++k) {
int x = i + dx[k], y = j + dy[k];
if (x < 1 || x > n || y < 1 || y > m) continue;
for (int l = 0; l <= id[i][j]; ++l)
for (int r = 0; r <= id[x][y]; ++r)
Add_edge(D(i, j) + l * n * m, D(x, y) + r * n * m, 1);
}
}
if (ans & 1) return cout << "N", 0;
cout << (ans / 2 == dinic() ? "Y" : "N") << "\n";
return 0;
}